Hamiltonian Mechanics Overflow#

Things in progress, not ready for the main notes yet.

WKB: Path Integral Formulation#

Warning

This section is still in progress…

In quantum mechanics, one can use the Feynman path-integral approach to construct the propagator (here expressed in terms of position-to-position transitions):

\[\begin{gather*} \newcommand{\S}{\mathcal{S}} U(q, t; q_0, t_0) = \int \mathcal{D}[q]\; \exp\left\{\frac{\I}{\hbar}S[q]\right\},\\ \psi(q, t) = \int U(q, t;q_0, t_0)\psi(q_0, t_0)\d{q_0}. \end{gather*}\]

where the integral is over all paths that start at $q(t_0) = q_0$ and end at $q(t) = q$, and $S[q]$ is the classical action

\[\begin{gather*} S[q] = \int_{t_0}^{t}\d{t}\; L(q, \dot{q}, t). \end{gather*}\]

Given an initial wavefunction $\psi(q_0, t_0)$, the wavefunction at time $t$ is:

\[\begin{gather*} \psi(q, t) = \int \d{q_0}\; U(q, t;q_0, t_0)\psi(q_0, t_0). \end{gather*}\]

the WKB approximation relies on the idea that classical trajectories where $S'[q_{\mathrm{cl}}] = 0$ – the famous principle of extremal action – dominate the propagator, and use the expansion of the action

\[\begin{gather*} S[q+\xi] = S[q] + S'[q]\cdot \xi + \frac{1}{2!}S''[q]\cdot\xi\xi + \frac{1}{3!}S'''[q]\cdot\xi\xi\xi + \cdots. \end{gather*}\]

The WKB approximation amount to considering all classical trajectories with appropriate boundary conditions, performing the path integral over $\xi$, and dropping terms of order $\order(\xi^3)$ and higher to obtain:

\[\begin{gather*} U_{WKB}(q, t; q_0, t_0) = \int \mathcal{D}[\xi]\; \exp\left\{\frac{\I}{\hbar}\left( S[q_{\mathrm{cl}}] + \frac{1}{2}S''[q_{\mathrm{cl}}]\cdot\xi\xi\right)\right\}\\ = \sqrt{\frac{-\partial^2 S / (2\pi \I \hbar)} {\partial q_{\mathrm{cl}}(t)\partial q_{\mathrm{cl}}(t_0)}} \exp\left\{\frac{\I}{\hbar}\S(q_{\mathrm{cl}}(t),t;q_{\mathrm{cl}}(t_0),t_0)\right\}, \end{gather*}\]

where $\S = \S(q,t;q_0,t_0)$ is the classical action with $q=q_{\mathrm{cl}}(t)$ and $q_0 = q_{\mathrm{cl}}(t_0)$ are the final and initial points of the classical trajectory. The key point here is that all of the information about the propagator in this approximation is contained in the classical action $\S(q,t;q_0,t_0)$, sometimes called Hamilton’s principal function.

Once the path integrals over $\xi$ have been done, everything is expressed in terms of the classical trajectory $q_{\mathrm{cl}}(t)$ and we shall drop the “cl” subscript in what follows.

(Note: if there are multiple trajectories that satisfy the boundary conditions, then they should be added, giving rise to quantum interference patterns.)

Example: Free Particle

As an example, consider a free particle with Hamiltonian $H = p^2/2m$. This has the following solution $x(t)$ and Hamilton’s principal function $\S$:

\[\begin{gather*} x(t) = x_0 + \frac{p_0}{m}(t-t_0), \qquad p_0 = m\frac{x-x_0}{t-t_0},\\ \S(x,t;x_0,t_0) = \frac{p_0^2}{2m}(t-t_0) = \frac{m(x-x_0)^2}{2(t-t_0)},\\ \frac{\partial^2 \S}{\partial x\partial x_0} = -\frac{m}{(t-t_0)}. \end{gather*}\]

Hence, the WKB propagator is:

\[\begin{gather*} U_{WKB}(x,t;x_0,t_0) = \sqrt{\frac{m}{2\pi \I \hbar (t-t_0)}}\exp\left\{ \frac{i}{\hbar}\frac{m(x-x_0)^2}{2(t-t_0)} \right\}. \end{gather*}\]

A quick check shows that this is exact:

\[\begin{gather*} U(x,t;x_0,t_0) = \braket{x|e^{\op{H}(t-t_0)/\I\hbar}|x_0} = \int \frac{\d{k}}{2\pi}\braket{x|k}e^{\hbar^2k^2(t-t_0)/(2m\I\hbar)}\braket{k|x_0}\\ = \int \frac{\d{k}}{2\pi}e^{\hbar^2k^2(t-t_0)/(2m\I\hbar) + \I k (x-x_0)}\\ = \sqrt{\frac{m}{2\pi \I\hbar (t-t_0)}}\exp\left\{ \frac{\I m(x-x_0)^2}{2\hbar (t-t_0)} \right\}. \end{gather*}\]

Extending this to higher dimensions, we have:

\[\begin{gather*} \vect{x}(t) = \vect{x}_0 + \frac{\vect{p}_0}{m}(t-t_0), \qquad \vect{p}_0 = m\frac{\vect{x}-\vect{x}_0}{t-t_0},\\ \S(\vect{x},t;\vect{x}_0,t_0) = \frac{m\norm{\vect{x}-\vect{x}_0}^2}{2(t-t_0)},\\ \frac{\partial^2 \S}{\partial x_{i}\partial [x_0]_{j}} = -\frac{m\delta_{ij}}{(t-t_0)}. \end{gather*}\]

Similar results can be obtained from the momentum-to-position transitions if the initial state is expressed in terms of momentum, however, in this case, since the boundary conditions are no longer the same, we must use a different form of $\S(x,t;p_0,t_0)$:

\[\begin{gather*} \S(q,t;p_0,t_0) = p_0q_0(q,t;p_0,t_0) + \S\Bigl(q,t;q_0(q,t;p_0,t_0),t_0\Bigr),\\ U_{WKB}(q, t; p_0, t_0) = \sqrt{\frac{\partial^2 \S}{\partial q_{\mathrm{cl}}(t)\partial p_{\mathrm{cl}}(t_0)}} \exp\left\{\frac{\I}{\hbar}\S(q_{\mathrm{cl}}(t),t;p_{\mathrm{cl}}(t_0);t_0)\right\}. \end{gather*}\]

Example: Free Particle continued

Changing variables, we now have:

\[\begin{gather*} \S(x,t;p_0,t_0) = p_0\left(x - \frac{p_0}{m}(t-t_0)\right) + \overbrace{\frac{p_0^2}{2m}(t-t_0)}^{S}\\ = p_0 x - \frac{p_0^2}{2m}(t-t_0), \qquad \frac{\partial^2 \S}{\partial x\partial p_0} = 1 \end{gather*}\]

Hence, the WKB propagator is:

\[\begin{gather*} U_{WKB}(x,t;p_0,t_0) = \exp\left\{ \frac{i}{\hbar}\left(xp_0 - \frac{p_0^2 (t-t_0)}{2m}\right)\right\}. \end{gather*}\]

A quick check shows that this is also exact, and confirms the need for the extra piece $q_0 p_0$ in $\S$:

\[\begin{gather*} U(x,t;p_0,t_0) = \braket{x|e^{\op{H}(t-t_0)/\I\hbar}|p_0} = \braket{x|p_0}e^{p_0^2(t-t_0)/(2m\I\hbar)}\\ = e^{\tfrac{\I}{\hbar} x p_0}e^{p_0^2(t-t_0)/(2m\I\hbar)}\\ = \exp\left\{\frac{\I}{\hbar}\left( x p_0 - \frac{p_0^2}{2m}(t-t_0) \right)\right\}. \end{gather*}\]

Examples#

Example: Falling Particle 1D

As a second example, consider a particle in free-fall with Hamiltonian $H = p^2/2m + mgz$. The classical problem is most easily solved with conservation of energy $E$:

\[\begin{align*} E &= \frac{p^2}{2m} + mgz = \frac{p_{0}^2}{2m} + mgz_0, \\ p &= \pm\sqrt{2mE - 2m^2gz},\\ &= \pm\sqrt{p_{0}^2 - 2m^2g(z-z_0)},\\ L(z) &= E - 2mgz. \end{align*}\]

Since our Hamiltonian is time-independent, the results will depend only on $t-t_0$ and we can choose $t_0=0$ without loss of generality. This has the following solution $z(t)$ and Hamilton’s principal function $\S$:

\[\begin{gather*} \newcommand{\t}{t} z(t) = z_0 + \frac{p_0}{m}\t - \frac{g}{2}\t^2,\\ \S(z,\t;z_0,t_0=0) = \int_{0}^{\t}\left(E - 2mgz_0 - 2gp_0\t + mg^2\t^2\right)\d{\t},\\ = E(z,\t;z_0)\t - 2mgz_0\t - gp_0(z,\t;z_0)\t^2 + \frac{mg^2\t^3}{3}. \end{gather*}\]

The appropriate functional dependence can be deduced by solving for $p_0(z,\t;z_0)$ and $E(z,\t;z_0)$:

\[\begin{align*} p_0(z,\t;z_0) &= m\frac{z-z_0}{\t} + \frac{mg\t}{2}, & \pdiff{p_0}{z_0} &= -\frac{m}{\t}, \\ E(z,\t;z_0) &= \frac{p_0^2(z,\t;z_0)}{2m} + mgz_0, & \pdiff{E}{z_0} &= -\frac{p_0}{\t} + mg. \end{align*}\]

From these, we can compute the action and various partials:

\[\begin{gather*} \S(z,\t;z_0,t_0=0) = m\left( \frac{(z-z_0)^2}{2\t} - \frac{g}{2}(z+z_0)\t - \frac{g^2\t^3}{24}\right),\\ \frac{\partial\S(z,\t;z_0,t_0=0)}{\partial z} = p = p_0(z, \t;z_0) - mg\t,\\ \frac{\partial\S(z,\t;z_0,t_0=0)}{\partial z_0} = -p_0\\ \frac{\partial^2\S(z,\t;z_0,t_0=0)}{\partial z\partial z_0} = -\frac{m}{\t}. \end{gather*}\]

Hence, the WKB propagator is:

\[\begin{gather*} U_{WKB}(z,t;z_0,t_0) = \sqrt{\frac{m}{2\pi \I \hbar (t-t_0)}}\\ \exp\left\{ \frac{\I m}{\hbar}\left( \frac{(z-z_0)^2}{2(t-t_0)} - \frac{g}{2}(z+z_0)(t-t_0) - \frac{g^2(t-t_0)^3}{24} \right) \right\}. \end{gather*}\]

Example: General Particle 1D

Slightly more general, we now consider a particle falling in an arbitrary time-independent potential $V(z) = mgz + \delta(z)$ where we will ultimately consider $\delta(z)$ to be small. Here, since energy is conserved, we immediately have:

\[\begin{gather*} E = \frac{p^2}{2m} + V(z) = \frac{p_{0}^2}{2m} + V(z_0), \\ p = \pm\sqrt{2mE - 2mV(z)},\\ L(z) = E - 2V(z). \end{gather*}\]

The trajectory $z(t)$ no longer has a closed form, but we can still express the action by changing variables $\d{z} = \dot{z}\d{t} = p\d{t}/m$:

\[\begin{gather*} \S(z;z_0;E) = m\int_{z_0}^{z}\frac{\Bigl(E - 2V(z)\Bigr)}{p(z)}\d{z}\\ = \sqrt{\frac{m}{2}}\int_{z_0}^{z}\frac{\Bigl(E - 2V(z)\Bigr)} {\pm\sqrt{E - V(z)}}\d{z}. \end{gather*}\]

This form has two complications. First, the sign of the denominator must be chosen appropriately to match the direction of motion. This is often clear from the physics, and so does not pose a fundamental problem. Second, this form of the action as an explicit function of either $S(z, p; z_0)$ or $S(z;z_0, p_0)$ since $E = E(z_0, p_0) = E(z, p)$ is conserved and a function of the initial or final coordinates.

A comment about the role of the conserved energy $E$ here. Note that if $E=0$, then the numerator and denominator both contain factors of $\sqrt{-V(z)}$ and can be combined. The presence of $E$ seems to spoil this, but, as is generally well known, in classical mechanics, only the relative value of the energy is physically significant. To make this explicit, we note that

\[\begin{gather*} V(z) = E - \frac{p^2}{2m}, \quad L(z) = E - 2V(z) = \frac{p^2}{m} - E,\\ \begin{aligned} S(z;z_0;E) &= -\int_{0}^{t}E\d{t} + \int_{z_0}^{z}\bigl(\pm p(z)\bigr)\d{z}\\ &= -Et \pm \int_{z_0}^{z}p(z)\d{z}. \end{aligned} \end{gather*}\]

The first term clearly does not affect the physics, and in quantum mechanics, corresponds to an overall global phase. This is exactly the effect of shifting the zero-energy level. The second term is a common form of the action, as an integral of a generalized momentum with respect to the corresponding coordinate. This form appears in the action-angle variable formulation for example.

To compute the normalization factor, we must perform the appropriate change of variables using the analogy of the Maxwell relations in thermodynamics using:

\[\begin{gather*} f(z, z_0, E) = t - t_0 = \int_{t_0}^{t}\d{t} = \int_{z_0}^{z}\frac{m}{p}\d{z}\\ = \sqrt{\frac{m}{2}}\int_{z_0}^{z}\frac{1}{\pm\sqrt{E - V(z)}}\d{z},\\ \pdiff{f(z, z_0, E)}{z} = \pm\frac{\sqrt{m/2}}{\sqrt{E-V(z)}}\\ \pdiff{f(z, z_0, E)}{z_0} = \mp\frac{\sqrt{m/2}}{\sqrt{E-V(z_0)}}\\ \pdiff{f(z, z_0, E)}{E} = \mp\int_{z_0}^{z}\frac{\sqrt{m/8}}{\sqrt{E - V(z)}^3}\d{z}. \end{gather*}\]

We must compute the partials holding this constant, so we have:

\[\begin{gather*} \pdiff{f(z, z_0, E)}{z} \d{z} + \pdiff{f(z, z_0, E)}{z_0} \d{z_0} + \pdiff{f(z, z_0, E)}{E} \d{E} = 0,\\ \d{E} = \frac{\pdiff{f(z, z_0, E)}{z}\d{z} + \pdiff{f(z, z_0, E)}{z_0}\d{z_0}} {\pdiff{f(z, z_0, E)}{E}}\\ = \frac{-2}{\int_{z_0}^{z}\frac{1}{\sqrt{E - V(z)}^3}\d{z}} \left(\frac{\d{z}}{\sqrt{E-V(z)}} - \frac{\d{z_0}}{\sqrt{E-V(z_0)}}\right). \end{gather*}\]

We can now compute the normalization factor. We take the first derivative using the well-known properties of Hamilton’s principle function

\[\begin{gather*} \frac{\partial^2 S(z,t;z_0,t_0)}{\partial z\partial z_0} = \frac{\partial p(z,t;z_0,t_0)}{\partial z_0} = -\frac{\partial p_0(z,t;z_0,t_0)}{\partial z} \end{gather*}\]

and then use the expression above for $p(z, E) = \pm \sqrt{2m\bigl(E-V(z)\bigr)}$ to compute:

\[\begin{gather*} \d{p} = \frac{\pm \sqrt{2m}}{\sqrt{E-V(z)}}\Bigl(\d{E} - V'(z) \d{z}\Bigr)\\ \pdiff{p(z, t;z_0, t_0)}{z_0} = \frac{\pm \sqrt{8m}}{\sqrt{\bigl(E-V(z)\bigr)\bigl(E-V(z_0)\bigr)}} \frac{1}{\int_{z_0}^{z}\frac{1}{\sqrt{E - V(z)}^3}\d{z}}. \end{gather*}\]

Example: Falling Particle 2D

As a second example, consider a particle in free-fall with Hamiltonian $H = (p_x^2 + p_z^2)/2m + mgz$. The classical problem is most easily solved with conservation of energy $E$ and momentum $p_x$:

\[\begin{gather*} E = \frac{p_x^2+p_z^2}{2m} + mgz = \frac{p_{x0}^2 + p_{z0}^2}{2m} + mgz_0, \\ p_z = \pm\sqrt{2mE - 2m^2gz - p_x^2}\\ = \pm\sqrt{p_{z0}^2 - 2m^2g(z-z_0) + (p_{x0}^2 - p_x^2)}\\ L(\vect{r},t) = E - 2mgz \end{gather*}\]

Again, time-independence allows us to set $t_0=0$ without loss of generality.

Incomplete

This has the following solution $\vect{r}(t)$ and Hamilton’s principal function $\S$:

\[\begin{gather*} \vect{r}(t) = \vect{r}_0 + \frac{\vect{p}_0}{m}(t - t_0) - \frac{g}{2}(t-t_0)^2\uvect{z},\\ p_0 = m\frac{x-x_0}{t-t_0},\\ \S(\vect{r},t;\vect{r}_0,t_0) = \frac{p_0^2}{2m}(t-t_0) = \frac{m(x-x_0)^2}{2(t-t_0)},\\ \frac{\partial^2 \S}{\partial x\partial x_0} = -\frac{m}{(t-t_0)}. \end{gather*}\]

Hence, the WKB propagator is:

\[\begin{gather*} U_{WKB}(x,t;x_0,t_0) = \sqrt{\frac{m}{2\pi \I \hbar (t-t_0)}}\exp\left\{ \frac{i}{\hbar}\frac{m(x-x_0)^2}{2(t-t_0)} \right\}. \end{gather*}\]

Changing variables, we now have:

\[\begin{gather*} \S(x,t;p_0,t_0) = xp_0 - \frac{p_0^2}{2m}(t-t_0),\qquad \frac{\partial^2 \S}{\partial x\partial p_0} = 1 \end{gather*}\]

Hence, the WKB propagator is:

\[\begin{gather*} U_{WKB}(x,t;p_0,t_0) = \exp\left\{ \frac{i}{\hbar}\left(xp_0 - \frac{p_0^2 (t-t_0)}{2m}\right)\right\}. \end{gather*}\]

A quick check shows that this is also exact, and confirms the need for the extra piece $q p_0$ in $\S$:

Example: General Particle 2D

As a second example, consider a particle in free-fall with Hamiltonian $H = (p_x^2 + p_z^2)/2m + V(x, z)$. The classical problem is most easily solved with conservation of energy $E$:

\[\begin{gather*} E = \frac{p_x^2+p_z^2}{2m} + V(x, z) = \frac{p_{x0}^2 + p_{z0}^2}{2m} + V(x_0, z_0), \\ p_z = \pm\sqrt{2mE - 2m^2gz - p_x^2}\\ = \pm\sqrt{p_{z0}^2 - 2m^2g(z-z_0) + (p_{x0}^2 - p_x^2)}\\ L(\vect{r},t) = E - 2mgz \end{gather*}\]

Again, time-independence allows us to set $t_0=0$ without loss of generality.

Incomplete

This has the following solution $\vect{r}(t)$ and Hamilton’s principal function $\S$:

Hence, the WKB propagator is:

\[\begin{gather*} U_{WKB}(x,t;x_0,t_0) = \sqrt{\frac{m}{2\pi \I \hbar (t-t_0)}}\exp\left\{ \frac{i}{\hbar}\frac{m(x-x_0)^2}{2(t-t_0)} \right\}. \end{gather*}\]

Changing variables, we now have:

\[\begin{gather*} \S(x,t;p_0,t_0) = xp_0 - \frac{p_0^2}{2m}(t-t_0),\qquad \frac{\partial^2 \S}{\partial x\partial p_0} = 1 \end{gather*}\]

Hence, the WKB propagator is:

\[\begin{gather*} U_{WKB}(x,t;p_0,t_0) = \exp\left\{ \frac{i}{\hbar}\left(xp_0 - \frac{p_0^2 (t-t_0)}{2m}\right)\right\}. \end{gather*}\]

A quick check shows that this is also exact, and confirms the need for the extra piece $q p_0$ in $\S$:

Example: Harmonic Oscillator

The harmonic oscillator has the following solution:

\[\begin{gather*} \newcommand{\t}{\tau} H(x, p) = \frac{p^2}{2m} + \frac{m\omega^2 x^2}{2}, \qquad \t= t-t_0\\ S(x, t; x_0, t_0) = \frac{\omega(xp - x_0p_0)}{2}\\ = \frac{m\omega}{2\sin\omega\t}\Bigl( (x^2+x_0^2)\cos\omega\t - 2xx_0\Bigr),\\ \frac{\partial^2 S}{\partial x \partial x_0} = -\frac{m\omega}{\sin\omega\tau},\qquad A = \sqrt{\frac{m\omega}{\sin\omega\tau}}\\ \begin{aligned} S' &\equiv \pdiff{S}{x} = p = \frac{m\omega}{\sin\omega\t}\Bigl(x\cos\omega\t - x_0\Bigr),\\ S'' &\equiv \pdiff[2]{S}{x} = m\omega\cot\omega\t,\\ \dot{A} &\equiv \pdiff{A}{t} %= \frac{-1}{2}\sqrt{\frac{m\omega^3\cos^2\omega \tau}{\sin^3\omega \tau}} = \frac{-\omega A \cot \omega \tau}{2} \end{aligned}\\ \frac{S''}{2m} + \frac{\dot{A}}{A} + \frac{A'S'}{mA} = \frac{\omega}{2}\cot\omega\t - \frac{\omega}{2}\cot\omega\t + 0 = 0. \end{gather*}\]

Traditional WKB#

The more traditional approach is to express the wavefunction as follows, then insert it into the Schrödinger equation:

\[\begin{gather*} \psi(x, t) = \exp\left\{\frac{i}{\hbar}W(x,t)\right\},\\ \left(\frac{(S')^2}{2m} - \frac{\I\hbar}{2m}W'' + V + \dot{W}\right)\psi = 0. \end{gather*}\]

Expanding $W$ in powers of $\hbar$, we have the following lowest two orders:

\[\begin{align*} &W(x,t) = S(x,t) - \I\hbar \log A + \order(\hbar^2),\\ \text{Order $\hbar^0$:}\quad &\frac{(S')^2}{2m} + V(x,t) + \dot{S} = H(x, S', t) + \dot{S} = 0\\ \text{Order $\hbar^1$:}\quad &\frac{S''}{2m} + \frac{A'S'}{mA} + \frac{\dot{A}}{A} = 0\\ &\psi_{WKB}(x, t) = A(x,t)\exp\left\{\frac{i}{\hbar}S(x,t)\right\}. \end{align*}\]

$\order(\hbar^0)$: Hamilton-Jacobi Equation#

The order $\hbar^0$ equation is the well-known Hamilton-Jacobi equation, which is satisfied by the classical action as a function of initial and final states.

Exercise

Prove that the classical action $S(x, t)$ satisfies the Hamilton-Jacobi equation:

\[\begin{gather*} S(x, t) = S(x,t;x_0,t_0) = \int_{t_0}^{t} L\Bigl(x(t), \dot{x}(t), t\Bigr) \d{t}, \end{gather*}\]

where $x(t)$ is a solution to the classical equations of motion with boundary conditions $x(t_0) = x_0$ and $x(t) = x$, as discussed above. I.e., show that

\[\begin{gather*} S' \equiv \pdiff{S(x, t; x_0, t_0)}{x} = p, \\ \dot{S} = \pdiff{S(x, t;x_0, t_0)}{t} = -H(x, S', t). \end{gather*}\]

$\order(\hbar^1)$: Continuity Equation#

The order $\hbar^1$ equation is the well-known continuity equation, expressing the conservation of particle number or probability. To see this, multiply through by $2A^2$:

\[\begin{gather*} \frac{A^2 S'' + 2S'A'A}{m} + 2\dot{A}A = 0\\ \left(\frac{S'}{m}A^2\right)' + \pdiff{A^2}{t} = 0. \end{gather*}\]

This is the 1-dimensional form of the familiar continuity equation, once we identify $S'$ as the momentum, and $A^2$ as the probability density:

\[\begin{gather*} \vect{\nabla}\cdot\vect{j} + \dot{n} = 0,\\ n \equiv A^2 = \abs{\psi}^2, \qquad \vect{v} \equiv \frac{\vect{p}}{m} = \frac{\vect{\nabla} S}{m}, \qquad \vect{j} = n\vect{v}. \end{gather*}\]

Exercise

Show that the order $\hbar^1$ equation is satisfied by

\[\begin{gather*} A(x, t) = \sqrt{-\frac{\partial^2 S(x, t; x_0, t_0) / (2\pi \I\hbar)} {\partial x\partial x_0}} \end{gather*}\]

as given by the path integral formulation. Note: this is only valid up to an overall constant giving the dependence of $A(x, t)$ on $x$ and $t$. One must adjust this overall constant to normalize the wavefunction.

Solution

Following the hint, we express $A^2 = \partial S'/\partial x_0 \equiv S'_{,x_0}$ (using Einstein’s notation). We are trying to show that:

\[\begin{gather*} \left(\frac{S' S'_{,x_0}}{m}\right)' + \pdiff{S'_{,x_0}}{t} = 0. \end{gather*}\]

We proceed using the linearity of the derivatives to rearrange the left-hand-side as:

\[\begin{gather*} \frac{1}{m}\pdiff{}{x}\Bigl( \overbrace{\pdiff{S}{x}\frac{\partial^2 S}{\partial x\partial x_0}} ^{\tfrac{1}{2}\partial (S')^2/\partial x_0} \Bigr) + \frac{\partial^2 \overbrace{\dot{S}}^{-H}}{\partial x\partial x_0} = \\ \frac{\partial^2}{\partial x_0 \partial x}\left( \frac{(S')^2}{2m} - H\right) = -\pdiff{V'(x, t)}{x_0} = 0, \end{gather*}\]

because the potential does not depend on the initial position $x_0$.

Exercise: Geometry

Note that since $S' = p$, the factor $A$ can be expressed as

\[\begin{gather*} n(x, t) \propto A^2(x, t) \propto \pdiff{p(x, t;x_0, t_0)}{x_0}. \end{gather*}\]

Explain this geometrically using Liouville’s theorem and the requirement that classical evolution conserves particle number

\[\begin{gather*} n_0(x_0, t_0)\d{x_0} = n(x, t)\d{x} \end{gather*}\]

for fixed initial condition $p_0(x_0)$.

Solution (Incomplete)

Liouville’s theorem tells us that evolution conserves phase-space volumes when considered as functions of $t$ and initial conditions $x_0$, and $p_0$ at time $t_0$:

\[\begin{gather*} x(t) = x(t; x_0, p_0, t_0), \qquad p(t) = p(t; x_0, p_0, t_0),\\ \frac{\partial (x, p)}{\partial (x_0, p_0)} = \pdiff{x}{x_0}\pdiff{p}{p_0} - \pdiff{x}{p_0}\pdiff{p}{x_0} = 1. \end{gather*}\]

\[\begin{gather*} \d{p} = \pdiff{p}{t}\d{t} + \pdiff{p}{x_0}\d{x_0} + \pdiff{p}{p_0}\d{p_0} + \pdiff{p}{t_0}\d{t_0},\\ \d{x} = \pdiff{x}{t}\d{t} + \pdiff{x}{x_0}\d{x_0} + \pdiff{x}{p_0}\d{p_0} + \pdiff{x}{t_0}\d{t_0}. \end{gather*}\]

To get $n(x, t)$ from the previous relationship, we hold $\d{x} = \d{t} = \d{t_0} = 0$.

\[\begin{gather*} \d{p} = \pdiff{p}{x_0}\d{x_0} + \pdiff{p}{p_0}\d{p_0},\\ \d{x} = 0 = \pdiff{x}{x_0}\d{x_0} + \pdiff{x}{p_0}\d{p_0},\\ \frac{\d{p_0}}{\d{x_0}} = -\frac{\pdiff{x}{x_0}}{\pdiff{x}{p_0}},\\ \pdiff{p(x, t;x_0, t_0)}{x_0} = \pdiff{p}{x_0} + \pdiff{p}{p_0}\frac{\d{p_0}}{\d{x_0}}\\ = \pdiff{p}{x_0} - \frac{\pdiff{x}{x_0}\pdiff{p}{p_0}}{\pdiff{x}{p_0}}\\ = \pdiff{p}{x_0} - \frac{1 + \pdiff{x}{p_0}\pdiff{p}{x_0}}{\pdiff{x}{p_0}}\\ = - \frac{1}{\pdiff{x}{p_0}}. \end{gather*}\]

Propagator#

We have been working with wavefunctions, but notice that the classical action is a function of both initial and final coordinates $S(x, t; x_0, t_0)$, but the initial coordinates are just parameters (they are not part of the Schrödinger equation). Since the Schrödinger equation is linear, we can form a solution as a linear combination of these, which allows us to use the full classical action to obtain an approximation to the quantum propagator, exactly mirroring the path integral approach:

\[\begin{gather*} \mat{U}_{WKB}(x, t;x_0, t_0) = \exp\left\{\frac{\I}{\hbar}W(x, t;x_0, t_0)\right\}\\ = A(x, t; x_0, t_0) \exp\left\{\frac{\I}{\hbar}S(x, t;x_0, t_0)\right\},\\ \psi_{WKB}(x, t) = \int \d{x_0}\; \mat{U}_{WKB}(x, t;x_0, t_0)\psi(x_0). \end{gather*}\]

The normalization needs to be checked, since the traditional WKB approach does not specify the magnitude of $A$, but, appropriately normalized, $\mat{U}_{WKB}$ is unitary.

Maxwell Relations#

To work with these expressions, we must compute the classical action for a particle which follows the classical trajectory, for example $S(q,t;q_0,t_0)$ for the position-to-position transitions, or $S(q,t;p_0,t_0)$ for the momentum-to-position transitions. We also need the various partial derivatives for the normalization factor and for additional analysis. Relating the partials of $S(q,t;q_0,t_0)$ to the partials of $S(q,t;p_0,t_0)$ follows the same process of deriving thermodynamic relationships such as the Maxwell relations.

Warm-up Exercise

Consider a function $f(q, q_0)$ and another variable $p_0(q, q_0)$. Show that the following hold:

\[\begin{align*} \newcommand{\mypd}[3]{\left.\pdiff{#1}{#2}\right|_{#3}} \mypd{f(q, p_0)}{p_0}{q} &= \frac{\mypd{f(q,q_0)}{q_0}{q}}{\mypd{p_0(q,q_0)}{q_0}{q}},\\ \mypd{f(q, p_0)}{q}{p_0} &= \mypd{f(q,q_0)}{q}{q_0} - \mypd{f(q,q_0)}{q_0}{q}\frac{\mypd{p_0(q,q_0)}{q}{q_0}}{\mypd{p_0(q,q_0)}{q_0}{q}}. \end{align*}\]

Check this by providing explicit forms for $f(q, q_0)$ and $p_0(q, q_0)$, then explicitly changing variables and differentiating.

Here we consider three different sets of variables $S(q,t;q_0,t_0)$, $S^{P}(q,t;p_0,t_0)$, and $S^{G}(q,t;q_0,p_0)$. To simplify the equations, we use the following notation for partials:

\[\begin{gather*} S_{,q} = \mypd{S(q,t;q_0,t_0)}{q}{t,q_0,t_0}, \qquad S^{P}_{,q} = \mypd{S^{P}(q,t;p_0,t_0)}{q}{t,p_0,t_0}, \\ S^{G}_{,q} = \mypd{S^{G}(q,t;q_0,p_0)}{t}{q,q_0,p_0}, \qquad p_{0,q} = \mypd{p_0(q,t;q_0,t_0)}{t}{t,q_0,t_0}, \qquad \text{etc.} \end{gather*}\]

I.e., the superscript denotes which set of variables is held fixed, and the subscript denotes which variable we differentiate. If there is no subscript Prove the following:

\[\begin{align*} S^{P}_{,p_0} &= \frac{S_{,q_0}}{p_{0,q_0}}, & S^{G}_{,p_0} &= \frac{S_{,t_0}}{p_{0,t_0}}, \\ S^{P}_{,q} &= S_{,q} - S_{,q_0}\frac{p_{0,q}}{p_{0,q_0}}, & S^{G}_{,q} &= S_{,q} - S_{,t_0}\frac{p_{0,q}}{p_{0,t_0}}, \\ S^{P}_{,t} &= S_{,t} - S_{,q_0}\frac{p_{0,t}}{p_{0,q_0}}, & S^{G}_{,t} &= S_{,t} - S_{,t_0}\frac{p_{0,t}}{p_{0,t_0}}, \\ S^{P}_{,t_0} &= S_{,t_0} - S_{,q_0}\frac{p_{0,t_0}}{p_{0,q_0}}, & S^{G}_{,q_0} &= S_{,p_0} - S_{,t_0}\frac{p_{0,q_0}}{p_{0,t_0}}. \end{align*}\]

To simplify these further, we need some mechanics. We shall work with $S(q,t;q_0,t_0)$ explicitly:

Exercise

Show that:

\[\begin{align*} S_{,q} &= p & S_{,t} &= -H & S_{,q_0} &= -p_0 & S_{,t_0} &= H_0. \end{align*}\]

Use the explicit form for the action, and then integrate by parts using the equations of motion to obtain simpler results. For some details, see [Houchmandzadeh, 2020].

These relationships allow us to derive formula similar to the Maxwell relations in thermodynamics. For example, using the fact that $p_0 = -S_{,q_0}$, $H_0 = S_{,t_0}$, and $\dot{p}_0 = -\partial H_0/\partial q_0$ from the Hamilton equations of motion, we have:

\[\begin{align*} p_{0,t_0} &= -\frac{\partial^2 S}{\partial q_0\partial t_0} = -\frac{\partial^2 S}{\partial t_0\partial q_0} = -\frac{\partial H_0}{\partial q_0} = \dot{p}_0. \end{align*}\]

Likewise, though less useful:

\[\begin{align*} p_{0,q} &= - p_{,q_0}, & p_{0,t} &= - H_{,q_0}, & p_{0,q_0} &= S_{,q_0,q_0}. \end{align*}\]

Falling Particle#

To check these relationships, we consider a particle falling in a gravitational field:

\[\begin{gather*} \newcommand{\t}{\tau} H(p, q) = \frac{p^2}{2m} + mgq, \qquad \t = (t-t_0),\\ \dot{p} = -\pdiff{H}{q} = -mg \quad \implies \quad p = p_0 - mg\t,\\ \dot{q} = \pdiff{H}{p} = \frac{p}{m} = \frac{p_0}{m} - g\t \quad \implies \quad q = q_0 + \frac{p_0}{m}\t - \frac{g}{2}\t^2. \end{gather*}\]

From this solution, we can construct the action $S(q_0, p_0, \t=t-t_0)$:

\[\begin{gather*} L = p\dot{q} - H = \frac{p^2}{2m} - mgq = \frac{p_0^2}{2m} - mgq_0 - 2gp_0\t + mg^2\t^2,\\ S(q_0, p_0, \t) = \left(\frac{p_0^2}{2m} - mgq_0\right)\t - gp_0\t^2 + \frac{mg^2}{3}\t^3. \end{gather*}\]

As a function of $S^{0}(t;p_0,q_0,t_0)$, this is not in any of the forms we considered above. To obtain those, we must eliminate one of the variables:

Independent Variables	Replacement	Name
$(q, t;q_0, t_0)$	$p_0(q, t;q_0, t_0)$	position-to-position
$(q, t;p_0, t_0)$	$q_0(q, t; p_0, t_0)$	momentum-to-position
$(q, t;q_0, p_0)$	$t_0(q, t; p_0, q_0)$	geometric optics

This requires inverting the equations to solve for $p_0$, $q_0$, or $t_0$ respectively. For the first two, we make use of the fact that the Hamiltonian (energy) is conserved $H(p_0, q_0) = H(p, q)$:

\[\begin{gather*} \frac{p^2}{2m} + mgq = \frac{p_0^2}{2m} mgq_0. \end{gather*}\]

For the last one, we need to solve the solution $q(\tau)$ for $\tau$. These give the following transformations:

\[\begin{align*} p_0(q, t, q_0, t_0) &= m\left(\frac{q-q_0}{\t} + \frac{g\t}{2}\right),\\ q_0(q, t, p_0, t_0) &= q + \frac{g\t^2}{2} - \frac{p_0\t}{m},\\ t_0(q, t, p_0, q_0) &= t - \frac{p_0}{mg} \mp \sqrt{\frac{p_0^2}{m^2g^2} -2 \frac{q - q_0}{g}} \end{align*}\]

In the last case, the appropriate branch must be chosen to meet the physical boundary conditions. Using these, we can express the action as:

\[\begin{align*} S(q,t;q_0,t_0) &= m\left(\frac{(q-q_0)^2}{2\t} - \frac{g(q+q_0)\t}{2} - \frac{g^2\t^3}{24} \right),\\ S^{P}(q,t;p_0,t_0) &= \left(\frac{p_0^2}{2m} - mgq\right)\t - \frac{mg^2\t^3}{6},\\ S^{G}(q,t;p_0,q_0) &= \frac{-p_0^3}{6m^2g} - p_0q_0 \mp \frac{\frac{p_0^2}{2m} + mg(2q+q_0)}{3g\sqrt{m/2}}\sqrt{\frac{p_0^2}{2m} - mg(q - q_0)},\\ &= \frac{2p_0(q-q_0) \mp \left(\frac{p_0^2}{2m} + mg(2q+q_0)\right)\t}{3}. \end{align*}\]

The expression for $S^{G}(q;p_0,q_0)$ is a bit messy, but notably does not depend on $t_0$ due to the time-invariance of the problem. However, to simplify expressions later, we present it in the second form where $\t$ should be replaced by

\[\begin{gather*} \t(q;p_0,q_0) = \frac{p_0}{mg} \pm \sqrt{\frac{p_0^2}{m^2g^2} -2 \frac{q - q_0}{g}}. \end{gather*}\]

Exercise

Check the relationships between the various partial derivatives of the actions. The following may be helpful:

\[\begin{gather*} p(q, t, q_0, t_0) = m\left(\frac{q-q_0}{\t} - \frac{g\t}{2}\right),\\ \begin{aligned} H = H_0 &= m\left(\frac{(q-q_0)^2}{2\t^2} + \frac{g^2\t^2}{8} + \frac{g(q+q_0)}{2}\right)\\ &= \frac{p_0^2}{2m} + mgq - gp_0\t + \frac{mg^2\t^2}{2}. \end{aligned} \end{gather*}\]

First with respect to the variables $(q, t;q_0, t_0)$:

\[\begin{align*} \pdiff{S}{q} &= p = m\left(\frac{q-q_0}{\t} - \frac{g\t}{2}\right),\\ \pdiff{S}{q_0} &= -p_0 = -m\left(\frac{q-q_0}{2\t} + \frac{g\t}{2}\right),\\ \pdiff{S}{t} &= -H = -m\left(\frac{(q-q_0)^2}{2\t^2} + \frac{g(q+q_0)}{2} + \frac{g^2\t^2}{8}\right),\\ \pdiff{S}{t_0} &= H_0 = m\left(\frac{(q-q_0)^2}{2\t^2} + \frac{g(q+q_0)}{2} + \frac{g^2\t^3}{8} \right). \end{align*}\]

Next, with respect to the variables $(q, t;p_0, t_0)$:

\[\begin{align*} \pdiff{S^{P}}{q} &= p - p_0 = -mg\t,\\ \pdiff{S^{P}}{t} &= -H + p_0\left(\frac{p_0}{m} -g\t\right) = \frac{p_0^2}{2m} - mgq - \frac{mg^2\t^2}{2} ,\\ \pdiff{S^{P}}{t_0} &= H_0 - p_0\left(\frac{p_0}{m} -g\t\right) =-\frac{p_0^2}{2m} + mgq + \frac{mg^2\t^2}{2},\\ \pdiff{S^{P}}{p_0} &= p_0\frac{\t}{m} =\frac{p_0}{m}\t. \end{align*}\]

Finally, with respect to the variables $(q, t;p_0, q_0)$:

\[\begin{align*} \pdiff{S^{G}}{q} &= p - \frac{H_0}{\frac{p_0}{m} - g\t} = ???,\\ \pdiff{S^{G}}{t} &= -H + H_0 = 0,\\ \pdiff{S^{G}}{q_0} &= -p_0 + \frac{H_0}{\frac{p_0}{m} - g\t} =???,\\ \pdiff{S^{G}}{p_0} &= H_0\left(\frac{p_0}{\t} - mg\right) = ???. \end{align*}\]

Atom Laser#

This section contains some notes about a continuous atom laser related to a research project I am working on. It might not make much sense if you are not familiar with the project. (Feel free to ask.)

Experimental Setup#

A continuous atom laser is formed by resonantly out-coupling atoms in a trapped BEC to a state that is not trapped, which falls under the influence of gravity and some additional potentials. (For more details, see Falling Atom Laser.)

The system can be modeled quite well by the following Schrödinger equation:

\[\begin{gather*} \I\hbar \partial_t \ket{\psi_a(t)} = \left(\frac{\op{p}_z^2}{2m} + V_a(\op{z}) - E_0\right)\ket{\psi_a(t)} + \Omega\ket{\psi_0} \end{gather*}\]

where $\ket{\psi_0}$ is the condensate wavefunction, and $E_0 = \mu - \hbar\omega$ is the chemical-potential of the condensate minus an energy shift due the frequency of the out-coupling, which can be used to shift where the atoms out-couple. (As discussed in Falling Atom Laser, efficient out-coupling happens within a small region around where $V_a(z) = E_0$.)

The out-coupled atoms fall under the influence of potential $V_a(z) \approx mgz$ where deviations occure due to small effects related to magnetic field gradients and optical “poky” potentials. As part of the experimental procedure, the atoms are subjected to $\pi$ or $\pi/2$ pulses, which create mixtures with another state $b$ which experiences a slightly different potential $V_b(z) = V_a(z) + \delta V(z) \approx V_a(z)$. These transitions occur about an unknown axis which we take to be $\hat{y}$ and have the following form:

\[\begin{gather*} \op{U}_{\theta} = e^{\I \theta \mat{\sigma}_y/2} = \cos\frac{\theta}{2}\mat{1} + \I\sin\frac{\theta}{2}\mat{\sigma}_y = \begin{pmatrix} \cos\frac{\theta}{2} & \sin\frac{\theta}{2}\\ -\sin\frac{\theta}{2} & \cos\frac{\theta}{2} \end{pmatrix}, \\ \op{U}_{\pi/2} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1\\ -1 & 1 \end{pmatrix}, \qquad \op{U}_{\pi} = \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}. \end{gather*}\]

Two procedures are used. Both start with a well-established atom laser in the quasi-stationary state $\ket{\psi_a}$ which we can, by shifting coordinates, take to satisfy:

\[\begin{gather*} \left(\frac{\op{p}_z^2}{2m} + V_a(\op{z})\right)\ket{\psi_a} = - \Omega e^{\I z_0 \op{p}_z/\hbar}\ket{\psi_0} \end{gather*}\]

The first interferometer applies a $\op{U}_{\pi/2}$ pulse at time $t_1$, then a second $\op{U}_{\pi/2}$ pulse at time $t_2 = t_1 + t_{\mathrm{wait}}$.
The second “spin-echo” interferometer first applies a $\op{U}_{\pi/2}$ pulse at time $t_1$, then a $\op{U}_{\pi}$ pulse at time $t_2$, and finally a $\op{U}_{\pi/2}$ pulse at time $t_3 = t_1 + t_{\mathrm{wait}}$.

Simplified WKB Analysis#

Heuristically, the two procedures give rise to the following propagation:

\[\begin{gather*} \mat{U}_{t_f, t_2} \underbrace{ \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1\\ -1 & 1\\ \end{pmatrix} }_{\mat{U}_{\pi/2}} \mat{U}_{t_2, t_1} \underbrace{ \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1\\ -1 & 1\\ \end{pmatrix} }_{\mat{U}_{\pi/2}} \mat{U}_{t_1, t_0} \begin{pmatrix} 1\\ 0 \end{pmatrix}\\ \mat{U}_{t_f, t_3} \underbrace{ \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1\\ -1 & 1\\ \end{pmatrix} }_{\mat{U}_{\pi/2}} \mat{U}_{t_3, t_2} \underbrace{ \begin{pmatrix} 0 & 1\\ -1 & 0\\ \end{pmatrix} }_{\mat{U}_{\pi}} \mat{U}_{t_2, t_1} \underbrace{ \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1\\ -1 & 1\\ \end{pmatrix} }_{\mat{U}_{\pi/2}} \mat{U}_{t_1, t_0} \begin{pmatrix} 1\\ 0 \end{pmatrix},\\ \mat{U}_{t_2, t_1} = \begin{pmatrix} e^{\I S^{a}_{12}/\hbar}\\ & e^{\I S^{b}_{12}/\hbar} \end{pmatrix}. \end{gather*}\]

Using the notation $A_{ij} = \exp(\I S^{a}_{ij}/\hbar)$ and $B_{ij} = \exp(\I S^{b}_{ij}/\hbar)$, we have:

Simple interferometer:

\[\begin{align*} \Psi(t_f) &\approx \frac{1}{2} \begin{pmatrix} A_{01}A_{12}A_{2f} - A_{01}B_{12}A_{2f}\\ -A_{01}A_{12}B_{2f} - A_{01}B_{12}B_{2f} \end{pmatrix}, \\ \begin{pmatrix} n_a(t_f)\\ n_b(t_f) \end{pmatrix} &\propto \begin{pmatrix} \abs{A_{12} - B_{12}}^2\\ \abs{A_{12} + B_{12}}^2 \end{pmatrix}. \end{align*}\]
Spin-echo interferometer:

\[\begin{align*} \Psi(t_f) &\approx \frac{1}{2} \begin{pmatrix} -A_{01}B_{12}A_{23}A_{3f} - A_{01}A_{12}B_{23}A_{3f}\\ A_{01}B_{12}A_{23}B_{3f} - A_{01}A_{12}B_{23}B_{3f} \end{pmatrix},\\ \begin{pmatrix} n_a(t_f)\\ n_b(t_f) \end{pmatrix} &\propto \begin{pmatrix} \abs{B_{12}A_{23} + A_{12}B_{23}}^2\\ \abs{B_{12}A_{23} - A_{12}B_{23}}^2 \end{pmatrix}. \end{align*}\]

To obtain an estimate for what we see, we now make some additional approximations:

We assume that the dominant contribution to the potentials $V_a(z) \approx V_b(z) \approx mgz$ is gravity, and that, consequently, the motion can be simply describe by free-fall. Under this approximation, we may view the particle as a two-component $SU(2)$-valued particle, whose components evolve on the Bloch sphere as the particle falls. To go beyond this, we must use some sort of multi-component WKB formalism, which is quite complicated.
We assume that the pulses $\op{U}_{\theta}$ are essentially instantaneous.
To obtain simple expressions, we shall also consider the limit where $t_{\mathrm{wait}} \rightarrow 0$, which we call the “impulse approximation”.

With these assumptions, we may consider a single particle falling from $z=0$ at some initial time $t_0$. The position and momentum are approximately:

\[\begin{gather*} q(t) \approx -\frac{g(t-t_0)^2}{2}, \qquad p(t) \approx -mg(t-t_0) = -m\sqrt{-2gq(t)}. \end{gather*}\]

The phase accumulates through the integral of the action

\[\begin{gather*} \delta S(z_1, z_2) = \int_{t_1}^{t_2}\mathcal{L}\d{t} = \int_{z_1}^{z_2}\frac{\mathcal{mL}}{p}\d{z},\qquad \mathcal{L} = E - 2 V(z),\\ V(z) = V_0(z)\mat{1} + \delta V(z) \mat{\sigma}_z = \begin{pmatrix} V_a(z)\\ & V_b(z) \end{pmatrix}. \end{gather*}\]

Under our approximations, we can take $E\approx 0$, so that

\[\begin{gather*} A_{ij}, B_{ij} \approx \exp\left( \frac{\I}{\hbar} \int_{z_i}^{z_j} \frac{2V_{a,b}(z)}{\sqrt{-2gz}} \right). \end{gather*}\]

Impulse approximation#

If the intervals $\delta t_{ij} = t_i - t_j$, then we can take the integrand to be constant over this interval, and use $z_j \approx z_i + \delta t_{ij} p_i/m$, which cancels the denominator:

\[\begin{gather*} A_{ij}, B_{ij} \approx \exp\left( \frac{\I}{\hbar} (z_i - z_j)\frac{2V_{a,b}(z_{ij})}{\sqrt{-2gz_{ij}}} \right) \approx \exp\left( -\frac{\I}{\hbar}\delta t_{ij}2V_{a,b}(z_{ij}) \right) \end{gather*}\]

Looking at the densities $n_{a,b}$ above, the simple interferometer measures the phase difference between $A_{12}$ and $B_{12}$, while the spin-echo interferometer measures the phase difference between $B_{12}A_{23}$ and $A_{12}B_{23}$:

Simple interferometer – measures contours of $\delta V(z)$:

\[\begin{gather*} \hbar\delta\phi \approx 2\delta t_{12}\bigl(V_{a}(z_{12})-V_{b}(z_{12})\bigr)\\ = 2 t_{w}\bigl(V_{a}(z_{12})-V_{b}(z_{12})\bigr) \end{gather*}\]
Spin-echo interferometer – measures contours of $\delta V'(z)$:

\[\begin{gather*} \hbar\delta\phi \approx 2\Bigl( \delta t_{12}\bigl(V_{b}(z_{12}) - V_{a}(z_{12})\bigr) - \delta t_{23}\bigl(V_{b}(z_{23}) - V_{a}(z_{23})\bigr) \Bigr),\\ \approx \frac{2t_{w}^2 p_{13}}{m} \Bigl(V_{a}'(z_{13}) - V_{b}'(z_{13})\Bigr) \end{gather*}\]

In the second expressions, we have taken $\delta t_{12} = \delta t_{23} = t_{w}$ and $z_{23} \approx z_{12} + t_w p_{13}/m$. This

Falling Gaussian#

Consider an initial state $\psi_0(z)$. The WKB approximation for the time evolution is:

\[\begin{gather*} \psi(z, t) = \int \d{z_0} I(z, z_0;t)\psi_0(z), \qquad I(z, z_0; t) = \sqrt{S_{,z,z_0}}e^{\tfrac{i}{\hbar}S(z, z_0;t)}, \end{gather*}\]

where the action $S(q,t;q_0,t_0)$ is computed over classical trajectories $q(0) = z_0$ and $q(t) = z$ – the first case above. We immediately have

\[\begin{gather*} S = m\left(\frac{(z-z_0)^2}{2 t} - \frac{g(z+z_0)t}{2} - \frac{g^2t^3}{24}\right),\\ S_{,z,z_0} = -\frac{m}{t}, \\ I(z, z_0; t) = \sqrt{-\frac{m}{t}}e^{\tfrac{i}{\hbar}S(z, z_0;t)},\\ \psi(z, t) = \int \d{z_0} I(z, z_0;t)\psi_0(z), \qquad \end{gather*}\]

\[\begin{gather*} S_{,z,z_0} = -\frac{m}{t}, \\ S^P_{,z,p_0} = 0, \\ S^{G}_{,z,z_0} = -\frac{m}{t}, \\ \end{gather*}\]

\[\begin{split} \mp \frac{1}{\sqrt{2m}} S^{G}_{,z} = \frac{\frac{p_0^2}{2m} - mg(2q - q_0)}{2\sqrt{\frac{p_0^2}{2m} - mg(q - q_0)}},\\ \mp \frac{1}{mg\sqrt{2m}} S^{G}_{,z,z_0} = \frac{\frac{p_0^2}{2m} + mgq_0}{4\sqrt{\frac{p_0^2}{2m} - mg(q - q_0)}^3}\\ S^{G}_{,z,z_0} = \mp \frac{m^2E V'(q)}{p^3} \end{split}\]

p^2 = 2m(E-V(q))

9999966626474851
999995478212815
9999236725503622
9601883448955634

../_images/f71bc006bfe762a89941266e003b05bfe8cd22370825c771fee80e7c17a899e5.png

Interference 1#

Now consider two streams of particles continuously injected at $z = 0$. The first steam falls without any external potential other than gravity, while the second experiences an additional potential $\lambda V(z)$. We choose our reference frame so that $H_0 = H = 0$. Then, for the first set of particles, we have:

\[\begin{gather*} z(t) = -g \frac{t^2}{2}, \qquad \dot{z} = -gt = -g\sqrt{\frac{-2z}{g}}, \qquad t = \sqrt{\frac{-2z}{g}},\\ S = -mgzt - \frac{mg^2t^3}{6} = \frac{mg^2t^3}{3} = \frac{mg^2}{3}\left(\frac{-2z}{g}\right)^{3/2}. \end{gather*}\]

If the potential $\lambda V(z)$ for the second species is small, we may use the Born approximation, under which the leading order correction to the action is:

\[\begin{gather*} S_a - S = - \lambda \int_{t_0}^{t}V\bigl(z(t)\bigr)\d{t} + \order(\lambda^2) = - \lambda \int_{0}^{z}\frac{V(z)}{\dot{z}}\d{z} + \order(\lambda^2)\\ = - \frac{\lambda}{\sqrt{2g}}\int_{0}^{z}\frac{V(z)}{\sqrt{-z}}\d{z} + \order(\lambda^2) \end{gather*}\]

from scipy.integrate import cumtrapz
micron = 1.0
mm = 1000*micron
meter = 1000*mm
sec = 1.0
amu = 1e-3
V0 = 85340

m = 87.0 * amu
g1 = 9.81 * meter/sec**2
g2 = 9.80 * meter/sec**2

x = np.linspace(-200, 200, 500)[:, None]
z = np.linspace(-400, 0, 502)[None, :]
z0 = -100

sigma = 20.0

def V(x, z):
    return V0 * np.exp(-((z-z0)**2+x**2)/2/sigma**2)
    
V1 = m * g1 * z + 0*x
V2 = m * g2 * z + 10*V(x, z)

dS1_dz = -np.sqrt(abs(-2*m*V1))
dS2_dz = -np.sqrt(abs(-2*m*V2))

S = cumtrapz((dS2_dz - dS1_dz)[:, ::-1], axis=1, initial=0)[:, ::-1]

Interference 2#

Now we consider a slightly different interference phenomenon, again with two streams injected with $p=0$ at $z=0$. The first falls in the potential $V(z)$ for all time, while the second experiences a different potential $V(z) + \Delta(z)$ for a short time interval between $t_1$ and $t_2 = t_1 + \delta_t$.

For the first particle, we have the usual:

\[ t_i - t_0^{a} = \int_0^{z_i} \frac{-m}{\sqrt{-2mV(Z)}}\d{Z}. \]

For the second particle, we must consider the three different time intervals:

\[\begin{gather*} t_i - t_0^{b} = \overbrace{ \int_0^{z_1} \frac{-m}{\sqrt{-2mV(Z)}}\d{Z} }^{t_1 - t_0^b}\\ + \overbrace{ \int_{z_1}^{z_2} \frac{-m}{\sqrt{-2m\Bigl(V(Z) + \Delta(Z) - \Delta(z_1)\Bigr)}}\d{Z} }^{t_2 - t_1}\\ + \overbrace{ \int_{z_2}^{z_i} \frac{-m}{\sqrt{-2m\Bigl(V(Z) - \Delta(z_1) + \Delta(z_2)\Bigr)}}\d{Z} }^{t_i - t_2}. \end{gather*}\]

If we assume that the motion is monotonic $p<0$, then we can use the geometric optics picture to re-express this in terms of view both particles falling through a conservative potential:

\[\begin{align*} V^a(z) &= V(z),\\ V^b(z) &= \begin{cases} V(z) & z_1 < z\\ V(z) + \Delta(z) - \Delta(z_1) & z_2 < z < z_1\\ V(z) + \Delta(z_2) - \Delta(z_1) & z < z_2. \end{cases} \end{align*}\]

Here we have exchanged $t_1$ and $t_2$ for $z_1$ and $z_2$. Of course, this must be done for each trajectory, but for a given trajectory, we can solve for the final momentum and action geometrically:

\[\begin{gather*} p^{a,b}(z) = -\sqrt{-2mV^{a,b}(z)},\qquad S^{a,b}(z) = \int_{0}^{z}p^{a,b}(z) \d{z}. \end{gather*}\]

The impulse approximation considers the limit $z_2 \rightarrow z_1$ so that

Something is wrong with $S$ here… should have an integral so $\Delta(z_1)$ not $\Delta'(z_1)$.

\[\begin{align*} V^b(z) &= V^a(z) + \delta_z \Delta'(z_1) \Theta(z_1 - z) + \order(\delta_z^2)\\ p^b(z) &= p^{a}(z) - \delta_z \Delta'(z_1) \Theta(z_1 - z) \frac{m}{p^{a}(z)} + \order(\delta_z^2)\\ S^b(z) &= S^{a}(z) - \delta_z \Delta'(z_1)\Theta(z_1 - z) \int_{z_1}^{z}\frac{m}{p^{a}(z)}\d{Z} +\order(\delta_z^2). \end{align*}\]

The only task now is to relate $z_1$ to

\[\begin{gather*} \delta_t = \delta_z\frac{m}{p^{a}(z_1)} + \order(\delta_z^2)\\ t_i - t_1 = \int_{z_1}^{z_i} \frac{m}{p^b(z)}\d{Z}\\ = \int_{z_1}^{z_i} \frac{m}{p^a(z)}\left( 1 + \delta_z \Delta'(z_1)\frac{m}{[p^{a}(Z)]^2} \right)\d{Z}+\order(\delta_z^2). \end{gather*}\]

To leading order, we can just solve for the reference particle to determine $z_1$:

\[\begin{gather*} t_i - t_1 = \int_{z_1}^{z_i} \frac{m}{p^a(Z)}\d{Z}. \end{gather*}\]

Then, the phase shift at $z_i$ will be:

\[\begin{gather*} S^b(z_i) - S^{a}(z_i) = -(t_i-t_1)\delta_z \Delta'(z_1)\Theta(z_1 - z) +\order(\delta_z^2). \end{gather*}\]

Hence, the phase shift directly maps the gradient of the potential at the position $z_1$.

From a geometric perspective, the motion of the second particle can be through of as described by a spatially dependent potential.

Hamiltonian Mechanics Overflow

Contents

Hamiltonian Mechanics Overflow#

WKB: Path Integral Formulation#

Examples#

Traditional WKB#

\(\order(\hbar^0)\): Hamilton-Jacobi Equation#

\(\order(\hbar^1)\): Continuity Equation#

Propagator#

Maxwell Relations#

Falling Particle#

Atom Laser#

Experimental Setup#

Simplified WKB Analysis#

Impulse approximation#

Falling Gaussian#

Interference 1#

Interference 2#

Independent Variables	Replacement	Name
\((q, t;q_0, t_0)\)	\(p_0(q, t;q_0, t_0)\)	position-to-position
\((q, t;p_0, t_0)\)	\(q_0(q, t; p_0, t_0)\)	momentum-to-position
\((q, t;q_0, p_0)\)	\(t_0(q, t; p_0, q_0)\)	geometric optics